Prove that $B = A^2 - 2A + 2I$ is an invertible matrix

Asked May 07 '23 at 19:20

Active May 07 '23 at 23:45

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If $A$ is a matrix such that $A^3 = 2I$, prove that $B = A^2 - 2A + 2I$ is an invertible matrix.

I tried to use $A^3 = 2I$ and turn it to $(A-I)(A^2 + A + I) = I$ but it didn't work. also I tried to suppose $B$ has inverse and solve the problem but it didn't work either.

edited May 07 '23 at 19:26

Rodrigo de Azevedo

23,223

asked May 07 '23 at 19:20

FATEMEH MOUSAVI

3

What have you tried so far – Fotis May 07 '23 at 19:21
Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Shaun May 07 '23 at 19:22
I tried to use $A^3$ = 2I and turn it to (A-I)($A^2$ + A + I) = I but it didn't work. also I tried to suppose B has inverte and solve the problem but it didn't work too. – FATEMEH MOUSAVI May 07 '23 at 19:23
@FATEMEHMOUSAVI Are you allowed to use diagonalization ? – TheSilverDoe May 07 '23 at 19:57
Yes, I think .. – FATEMEH MOUSAVI May 07 '23 at 20:00
@FATEMEHMOUSAVI I will write an answer whether the question reopens. – TheSilverDoe May 07 '23 at 20:09
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Take a look at the question Show that $B$ is invertible if $B=A^2-2A+2I$ and $A^3=2I$. I like the miracle method usage, and the dimension condition isn't used in any of the answers. – Sarvesh Ravichandran Iyer May 08 '23 at 00:01
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The systematic way is to compute polynomial gcd: see https://www.wolframalpha.com/input?i=simplify+PolynomialExtendedGCD%5Bx%5E2-2x%2B2%2Cx%5E3-2%2Cx%5D – lhf May 08 '23 at 00:06

Prove that $B = A^2 - 2A + 2I$ is an invertible matrix

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